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Introduction to Bases in Banach Spaces

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Introduction to Bases in Banach Spaces Introduction to Bases in Banach Spaces Matt Daws June 5, 2005 Abstract We introduce the notion of Schauder bases in Banach spaces, aiming to be able to give a statement of, and make sense of, the Gowers Dichotomy Theorem. Contents 1 Introduction 1 2 Schauder bases 2 3 Basic sequences 4 4 Unconditional bases 6 5 The Gowers Dichotomy Theorem 9 1 Introduction When working with finite dimensional vector spaces, it is often convenient to take a basis, and then work with the co-ordinate system which this basis gives us. In an infinite dimensional vector space, the axiom of choice shows that we can still find a basis: however, because it is now infinite, its use is often less. This is especially true in a Banach space, because such an algebraic basis takes no account of the extra structure induced by the norm. This leads to the notion of a Schauder basis. Let E be a Banach space and let (en) be a sequence of vectors in E. Then (en) is a Schauder basis (or, from now on, simply a basis) if every x ∈ E admits an expansion of the form ∞ X x = xnen, n=1 for some unique sequence of scalars (xn). Note. As someone interested in algebraic questions, I always work over the complex numbers. However, it is common (for simplicity) to work with real scalars in Banach space theory. Often, a similar result for complex numbers follows easily, sometimes we have to work somewhat harder, and occasionally, the result is true only for real scalars. We shall try to give proofs which work with either choice of scalar, but occasionally the reader should assume the result is only certainly known for real scalars. 1 p p Example 1.1. Let E = l for 1 ≤ p < ∞, or E = c0. Note that some authors write lp, ` , or `p instead. For n ≥ 1, let en ∈ E be the sequence which is 0 except with a 1 in the nth position. Then (en) is a Schauder basis for E, called the “standard unit-vector basis” of E. Example 1.2. Let H be a separable Hilbert space with an orthonormal basis (en). Then (en) is a basis (in fact, an unconditional basis: see later) for H. 2 Schauder bases We first give some general results on Schauder bases. Let E be a Banach space with a basis (en). Clearly, we are free to multiply each en by a non-zero scalar without changing the basis property. Hence we may assume that (and will henceforth do so) (en) is a normalised basis, that is, kenk = 1 for each n. Then, for x ∈ E, we can uniquely write P x = xnen, and hence define N X kxk0 = sup xnen . N≥1 n=1 PN This is well-defined, as n=1 xnen → x as N → ∞, and every convergent sequence in a Banach space is clearly bounded. We also easily see that kxk ≤ kxk0 for each x ∈ E, and that kxk0 = 0 if and only if x = 0. Theorem 2.1. Let E be a Banach space with a basis (en), and let k · k0 be as defined above. Then k · k and k · k0 are equivalent norms on E. Proof. Let F be the normed space which is E together with the norm k · k0 (it is easily checked that F is indeed a normed-space). Let ι : F → E be the formal inclusion map. Then ι is norm-decreasing, and is a bijection. Suppose that F is Banach space, that is, F is complete. Then the Open Mapping Theorem implies that ι has a continuous inverse, which implies that k · k0 is equivalent to k · k. P∞ So, we wish to show that F is complete. Firstly, given x = n=1 xnen, and m ≥ 1, we notice that m m−1 −1 −1 X X |xm| = kxmemkkemk = kemk xnen − xnen n=1 n=1 m m−1 −1 X X −1 ≤ kemk xnen + xnen ≤ 2kemk kxk0, n=1 n=1 by the definition of k · k0. Now let (xn) be a Cauchy-sequence in F , and let ∞ X xn = xn,mem (n ≥ 1). m=1 Then, for > 0, there exists N so that for r, s ≥ N, N X kxr − xsk0 = sup (xr,m − xs,m)em < . N≥1 m=1 By the calculation above, we hence see that for each m ≥ 1, the sequence of scalars ∞ P (xn,m)n=1 is a Cauchy-sequence, with limit ym say. We now show that y = ymem 2 converges in F . As k · k0 ≥ k · k, this implies that the sum converges in E, and hence that y exists, as E is complete. Towards this end, let n ≥ N, so that for each N ≥ 1, we have that N N N X X X (ym − xn,m)em = lim (xr,m − xn,m)em = lim (xr,m − xn,m)em ≤ , r→∞ r→∞ m=1 m=1 m=1 by the definition of N. Notice that this shows that y is indeed the limit of (xn), supposing P∞ of course that y exists. Then m=1 xn,mem converges in E, so there exists M such that Ps if M ≤ r < s, then m=r xn,mem < . Thus we have that s min(t,s) X X ymem = sup ymem 0 t≥r m=r m=r t t X X ≤ sup (ym − xn,m)em + xn,mem r≤t≤s m=r m=r t r−1 X X < sup (ym − xn,m)em + (ym − xn,m)em + ≤ 3, r≤t≤s m=1 m=1 P so that m ymem does converge in F . We can now give a “finite” characterisation of a basis. Theorem 2.2. Let E be Banach space, and let (en) be a sequence in E. Then (en) is a basis for E if and only if: 1. each en is non-zero; 2. the linear space of (en) is dense in E; 3. there exists a constant K such that for every sequence of scalars (xn), and each N < M, we have that N M X X xnen ≤ K xnen . n=1 n=1 Proof. Suppose that (en) is a basis, and form the norm k · k0, as above, so that by the above theorem, there exists K such that kxk ≤ kxk0 ≤ Kkxk (x ∈ E). P∞ Clearly we have condition (1) by the uniqueness of the expansion of 0 = n=1 0en; condition (2) is trivial. Then, for (3), let (yn) be the sequence of scalars defined by yn = xn for n ≤ M, and yn = 0 otherwise. Then we see that N N ∞ ∞ M X X X X X xnen = ynen ≤ ynen ≤ K ynen = K xnen , 0 n=1 n=1 n=1 n=1 n=1 as required. Conversely, let F be the linear span of (en), a dense subspace of E. For each n ≥ 1, define a linear map Pn : F → F by N n N X X X Pn xkek = xkek N ≥ n, xkek ∈ F . k=1 k=1 k=1 3 By condition (3), we see that kPnk ≤ K. It is then clear that Pn is a bounded projection on F , and that Pn thus extends to a bounded projection on E. For each n ≥ 1, define a ∗ linear functional en on F by the formula ∗ en(x)en = Pn(x) − Pn−1(x)(x ∈ F ), ∗ ∗ where P0 = 0. Then en is well-defined upon F (by condition (1)) and kenk ≤ 2K, so that ∗ en extends by continuity to a bounded linear functional on E. It is clear that n X ∗ Pn(x) = ek(x)ek (x ∈ F ), k=1 so by continuity, this formula holds for x ∈ E as well. Now let x ∈ E. We claim that ∞ X ∗ x = en(x)en. n=1 Let (xn) be a sequence in F which converges to x. Then, for > 0, let M ≥ 1 be such that kx − xM k < , and let N ≥ 1 be sufficiently large so that PN (xM ) = xM , which we can do, as xM ∈ F . Then we see that N X ∗ en(x)en − x = kPN (x) − xk n=1 ≤ kPN (x) − PN (xM )k + kPN (xM ) − xM k + kXM − xk ≤ Kkx − xM k + kxM − xk < (K + 1). Finally, we note that if ∞ ∞ X X ∗ x = xnen = en(x)en, n=1 n=1 ∗ ∗ then applying eN , we conclude that xN = eN (x) for each N, so that such an expansion is unique. The smallest constant which can arise in (3) is the basis constant of the basis (en). Theorem 2.1 shows that we can always renorm E to give a basis on 1. Such bases are called monotonic. It is immediate that a Banach space with a basis is separable, so that, for example, l∞ does not have a basis. Most common separable Banach spaces do have bases, although they are often non-obvious to find. For a while, it was thought that all separable Banach spaces would have a basis, but Enflo produced a counter-example in [Enflo, 1973]. There is, however, a weaker notion where we do have a positive result. 3 Basic sequences Let E be a Banach space, and let (xn) a sequence in E such that (xn) is a basis for its closed linear span. Then (xn) is a basic sequence in E. It is easy to show that every Banach space has a basic sequence; that is, every Banach space E contains a closed, infinite-dimensional subspace F with a basis. 4 Lemma 3.1. Let E be an infinite-dimensional Banach space, let F be a finite-dimensional subspace of E, and let > 0.
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